There are 16 components in a 4x4 matrix, but because the matrix is symmetrical the 6 components on the upper diagonal are equal to the 6 components on the lower diagonal. This makes 10 unique components (4 diagonal + 6 above it) and 16 total components.

Is gravity a result of the curvature of spacetime, or is it the curvature of spacetime a result of gravity?

Both.

Einstein's equation,

[tex]G = 8 \pi T[/tex]

is just that - an equation. [itex]G[/itex] on the the left refers to geometry, while [itex]T[/itex] on the right refers to movement and distrbution of matter/energy.

When solving Eintein's equation, physicists: sometimes consider a plausible distribution [itex]T[/itex] of matter, and try to find the geometry [itex]G[/itex] that produces this distribution; at other times consider a particular geometry [itex]G[/itex] and try to find the distribution of matter [itex]T[/itex] that produces this geometry.

John Wheeler said "Spacetime tells matter how to move. Matter tells spacetime how to curve." I'm quite sure that he meant "Spacetime curvature tells matter how to move. Matter tells spacetime how to curve."

YES U ULESS U FIND A GRAVITY AROUND A REGION U CAN'T BIND THE TIME OVER THERE BECAUSE AS U OBSERVE THE IS TIME LAGING ON MOON COMAPRED TO EARTH BECAUSE THERE IS A GAVITY LESS COMAPRED TO EARTH SO GRAVIT6Y IS THE CURAVATURE OF SPACE TIME AND VICE VERSA CAN'T BE RIGHT THING TO BE PRETCTED I HOPE THIS WOULD SATISFY .

If you turn off your caps locks and explain yourself slightly better someone might be able to give a decent response to your post.

Suffice to say, if a region of space is experiencing a graviational field, then there is also a warping of time's passage too. I'm not sure from your post if you're arguing with that, but it's the vague impression I get.

Yes but the metric is NOT a 4x4 matrix, it is a 2nd rank antisymmetric tensor. If you write it out in some basis it'll look like a 4x4 matrix, but fundamentally a 2nd rank antisymmetric tensor defined in 4 dimensions has 10 degrees of freedom, and therefore 10 components uniquely determine it.

Yes but the metric is NOT a 4x4 matrix, it is a 2nd rank antisymmetric tensor. If you write it out in some basis it'll look like a 4x4 matrix, but fundamentally a 2nd rank antisymmetric tensor defined in 4 dimensions has 10 degrees of freedom, and therefore 10 components uniquely determine it.

I think you meant a symmetric rank two tensor. That has ten independent components. Using a basis rep, they are the four components on the diagonal [tex] g_{\mu\mu}[/tex] and the six "above the diagonal" [tex]g_{\mu\nu}, \mu > \nu[/tex]. The six "below the diagonal" are just equal to the aboves [tex]g_{\mu\nu} = g_{\nu\mu}[/tex] in the symmetric tensor. These 4 and 6 and 6 make up the whole 16 components of the general rank 2 tensor in four dimensions.

In an antisymmetric tensor the diagonal elements always vanish and in a rank two tensor there are only six independent components.

But the stress-energy tensor still has 10 components, not 16. Independent or not. I could easily invent an object that had more components based on the 10 already; still that would not justify saying it had any more than the 10 independent ones.

But the stress-energy tensor still has 10 components, not 16. Independent or not. I could easily invent an object that had more components based on the 10 already; still that would not justify saying it had any more than the 10 independent ones.

The values of the components of a tensor is dependent on the specific coordinate system chosen. However, the number of components is not so dependent. The number of components depends on the dimension of the space - and there are 16 non-unique components for a 4-dimensional rank-2 tensor.

Part of the difference may be the specific approach used - I am using in this instance what the Wikipedia calls the "classic" approach to tensors.

There are equivalent approaches to visualizing and working with tensors; that the content is actually the same may only become apparent with some familiarity with the material.

* The classical approach

The classical approach views tensors as multidimensional arrays that are n-dimensional generalizations of scalars, 1-dimensional vectors and 2-dimensional matrices. The "components" of the tensor are the indices of the array. This idea can then be further generalized to tensor fields, where the elements of the tensor are functions, or even differentials.

The tensor field theory can roughly be viewed, in this approach, as a further extension of the idea of the Jacobian.

* The modern approach

The modern (component-free) approach views tensors initially as abstract objects, expressing some definite type of multi-linear concept. Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra. This treatment has largely replaced the component-based treatment for advanced study, in the way that the more modern component-free treatment of vectors replaces the traditional component-based treatment after the component-based treatment has been used to provide an elementary motivation for the concept of a vector. You could say that the slogan is 'tensors are elements of some tensor space'.

The emphasis on the term "multidimensional arrays" was added by me to point out the section of the defintion which justified my usage.

In any event, regardless of whether or not you are using the classic or abstract approach to tensors, the number of components is fixed by the dimensionality of the underlying vector space. Thus a rank-1 4d tensor requires exactly 4 components, no more, no less, and a rank-2 4d tensor requires exactly 16.

As we have discussed, there are 10 unique components in a symmetric rank 2 4d tensor, but without the qualifier "unique", one should say that there are 16 components, not 10.

I do agree with you, and I am sure we are both aware of exactly what we mean.

But ultimately, a tensor is a specific mathematical (in fact geometric) object whose components (and basis vectors) transform in a specific way. And the stress-energy tensor, which is something that exists regardless of a co-ordinate basis is spcified by 10 different components.

Anyway, there are, of course, two equivalent ways of looking at tensors, as you point out, and maybe we'll never agree. One of the reasons I am having this discussion is primarily to ensure that my viewpoint is not incorrect, and if it is, then to have it corrected. After all, I am yet to have any formal teaching on the topic.

Is gravity a result of the curvature of spacetime, or is it the curvature of spacetime a result of gravity? Or something else? Thank you.

You can define gravity to suit your assumptions you have about it. But I will address only Einstein's definition/view of gravity. Spacetime curvature is another term for tidal forces and where there is spacetime curvature there is a gravitational field. However the converse need not be true. It is quite possible to have a gravitational field without spacetime curvature. In fact the very first gravitational field that was of concern to Einstein was one which had zero spacetime curvature. That gravitational field is a uniform gravitational field. Hence the Equivalence Principle which states

A uniformly accelerating frame of reference is identical to a uniform gravitational field.

Einstein stated his objection to the "Gravity = curvature" interpretation in a letter to Max Von Laue.